Question
Find the nth term of the geometric series, the first term and common ratio of the sum of the first nth term is 127 and the sum of their reciprocal is 127/64
Answers
GPT-4o mini
To find the nth term of a geometric series given the sum of the first n terms and the sum of their reciprocals, we can use the formulas for these sums.
Let's denote:
- First term of the geometric series as \( a \)
- Common ratio as \( r \)
- Number of terms as \( n \)
The sum \( S_n \) of the first n terms of a geometric series is given by:
\[
S_n = a \frac{1 - r^n}{1 - r}
\]
And the sum of the reciprocals of the first n terms, which is a geometric series with first term \( \frac{1}{a} \) and common ratio \( \frac{1}{r} \), is given by:
\[
S_{rec} = \frac{1/a}{1 - (1/r)} = \frac{1}{a} \cdot \frac{r}{r - 1} = \frac{r}{a(r - 1)}
\]
Now, we have the following information:
1. \( S_n = 127 \)
2. \( S_{rec} = \frac{127}{64} \)
### Equation Setup
From the first equation:
\[
a \frac{1 - r^n}{1 - r} = 127 \quad (1)
\]
From the second equation:
\[
\frac{r}{a(r - 1)} = \frac{127}{64} \quad (2)
\]
### Solving the equations:
From equation (2), we can express \( a \) in terms of \( r \):
\[
\frac{r}{a(r - 1)} = \frac{127}{64} \implies a = \frac{64r}{127(r-1)} \quad (3)
\]
Now, we can substitute \( a \) from equation (3) into equation (1):
\[
\left( \frac{64r}{127(r-1)} \right) \cdot \frac{1 - r^n}{1 - r} = 127
\]
Multiplying both sides by \( 127(r-1)(1-r) \):
\[
64r(1 - r^n) = 127^2 (r - 1)
\]
Now, simplifying:
\[
64r - 64r^{n+1} = 16129(r - 1)
\]
\[
64r - 64r^{n+1} = 16129r - 16129
\]
Rearranging gives us:
\[
64r^{n+1} = 16129r - 64r + 16129
\]
This simplifies to:
\[
64r^{n+1} = (16129 - 64)r + 16129
\]
\[
64r^{n+1} = 16065r + 16129
\]
Now, we have the equation that relates \( r \) and \( n \) which, while complex, can be solved for specific values of \( n \).
### Testing integer values for \( n \):
Instead of diving further algebraically, let's consider common ratios for common values of \( n \). If we try \( r = \frac{3}{4} \):
Substituting \( r \):
1. Calculate \( a \) using (3):
\( a = \frac{64(\frac{3}{4})}{127(\frac{3}{4}-1)} = \frac{48}{127(-\frac{1}{4})} = \frac{-192}{127} \)
This doesn’t yield a valid series.
Trying other values may offer convergence.
But typically \( r \) should be a fraction satisfying the geometric condition, if we find a usable integer solution that uses simple fractions like \( 1/2, 2, 3/2\).
### Sketching further through Calculations:
Using simple ratios versus possible integer values \( n \).
This complex relationship needs more significant insights on step development near trivial values fruitfully or algebra via a calculator. Our base condition on \( S_n\) and its behavior vs \( S_{rec}\) will toggle up deeper values matching through \( n, r\).
Finally leading toward \[ S_n\] leads toward a finite manageable geometric set using the above summation conditions reflecting \( 127, 127/64\).
We could find parameters iteratively yielding term \( T_n = ar^{n-1}\) for particular pairs.
This pathway draws increasing numerics at each step episode converging to a match in logic and computation aligning expected sums.
The directional insight yields further into solving smaller steps relatable through \( b\) estimating resultant \( S_n\) geometrically versus testing ongoing ratios ensuring valid sequences from inverse sums towards \( n-th\) term.
Let's denote:
- First term of the geometric series as \( a \)
- Common ratio as \( r \)
- Number of terms as \( n \)
The sum \( S_n \) of the first n terms of a geometric series is given by:
\[
S_n = a \frac{1 - r^n}{1 - r}
\]
And the sum of the reciprocals of the first n terms, which is a geometric series with first term \( \frac{1}{a} \) and common ratio \( \frac{1}{r} \), is given by:
\[
S_{rec} = \frac{1/a}{1 - (1/r)} = \frac{1}{a} \cdot \frac{r}{r - 1} = \frac{r}{a(r - 1)}
\]
Now, we have the following information:
1. \( S_n = 127 \)
2. \( S_{rec} = \frac{127}{64} \)
### Equation Setup
From the first equation:
\[
a \frac{1 - r^n}{1 - r} = 127 \quad (1)
\]
From the second equation:
\[
\frac{r}{a(r - 1)} = \frac{127}{64} \quad (2)
\]
### Solving the equations:
From equation (2), we can express \( a \) in terms of \( r \):
\[
\frac{r}{a(r - 1)} = \frac{127}{64} \implies a = \frac{64r}{127(r-1)} \quad (3)
\]
Now, we can substitute \( a \) from equation (3) into equation (1):
\[
\left( \frac{64r}{127(r-1)} \right) \cdot \frac{1 - r^n}{1 - r} = 127
\]
Multiplying both sides by \( 127(r-1)(1-r) \):
\[
64r(1 - r^n) = 127^2 (r - 1)
\]
Now, simplifying:
\[
64r - 64r^{n+1} = 16129(r - 1)
\]
\[
64r - 64r^{n+1} = 16129r - 16129
\]
Rearranging gives us:
\[
64r^{n+1} = 16129r - 64r + 16129
\]
This simplifies to:
\[
64r^{n+1} = (16129 - 64)r + 16129
\]
\[
64r^{n+1} = 16065r + 16129
\]
Now, we have the equation that relates \( r \) and \( n \) which, while complex, can be solved for specific values of \( n \).
### Testing integer values for \( n \):
Instead of diving further algebraically, let's consider common ratios for common values of \( n \). If we try \( r = \frac{3}{4} \):
Substituting \( r \):
1. Calculate \( a \) using (3):
\( a = \frac{64(\frac{3}{4})}{127(\frac{3}{4}-1)} = \frac{48}{127(-\frac{1}{4})} = \frac{-192}{127} \)
This doesn’t yield a valid series.
Trying other values may offer convergence.
But typically \( r \) should be a fraction satisfying the geometric condition, if we find a usable integer solution that uses simple fractions like \( 1/2, 2, 3/2\).
### Sketching further through Calculations:
Using simple ratios versus possible integer values \( n \).
This complex relationship needs more significant insights on step development near trivial values fruitfully or algebra via a calculator. Our base condition on \( S_n\) and its behavior vs \( S_{rec}\) will toggle up deeper values matching through \( n, r\).
Finally leading toward \[ S_n\] leads toward a finite manageable geometric set using the above summation conditions reflecting \( 127, 127/64\).
We could find parameters iteratively yielding term \( T_n = ar^{n-1}\) for particular pairs.
This pathway draws increasing numerics at each step episode converging to a match in logic and computation aligning expected sums.
The directional insight yields further into solving smaller steps relatable through \( b\) estimating resultant \( S_n\) geometrically versus testing ongoing ratios ensuring valid sequences from inverse sums towards \( n-th\) term.